A182533 A symmetrical triangle. Read by rows: T(n,k) = 2*C(n-2,k-1) - C(n-2,k) - C(n-2,k-2), n >= 2, 0 <= k <= n, with T(0,0) = 0, T(1,0) = T(1,1) = 1.

0, 1, 1, -1, 2, -1, -1, 1, 1, -1, -1, 0, 2, 0, -1, -1, -1, 2, 2, -1, -1, -1, -2, 1, 4, 1, -2, -1, -1, -3, -1, 5, 5, -1, -3, -1, -1, -4, -4, 4, 10, 4, -4, -4, -1, -1, -5, -8, 0, 14, 14, 0, -8, -5, -1, -1, -6, -13, -8, 14, 28, 14, -8, -13, -6, -1, -1, -7, -19, -21, 6, 42, 42, 6, -21, -19, -7, -1

Offset: 1

Author

Alzhekeyev Ascar M, May 04 2012

Keywords

• sign
• tabl

Examples

			Triangle begins
   0;
   1,  1;
  -1,  2, -1;
  -1,  1,  1, -1;
  -1,  0,  2,  0, -1;
  -1, -1,  2,  2, -1, -1;
  -1, -2,  1,  4,  1, -2, -1;
  -1, -3, -1,  5,  5, -1, -3, -1;
  -1, -4, -4,  4, 10,  4, -4, -4, -1;
		

Links

• Muniru A Asiru, Table of n, a(n) for n = 1..5151 (rows n=1..100)
• Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, Graça Tomaz, Combinatorial Identities Associated with a Multidimensional Polynomial Sequence, J. Int. Seq., Vol. 21 (2018), Article 18.7.4.
• Graphical representation

Crossrefs

Cf. A000108, A120009, A137742.

Programs

GAP

T:=Concatenation([0,1,1],Flat(List([2..11],n->List([0..n],k->2*Binomial(n-2,k-1)-Binomial(n-2,k)-Binomial(n-2,k-2))))); # Muniru A Asiru, Nov 29 2018

Mathematica

Flatten[Join[{{0}, {1, 1}}, Table[2*Binomial[n - 2, k - 1] - Binomial[n - 2, k] - Binomial[n - 2, k - 2], {n, 2, 15}, {k, 0, n}]]] (* T. D. Noe, May 08 2012 *)

Formula

T(n,k) = 2*C(n-2,k-1) - C(n-2,k) - C(n-2,k-2).

Row sums = 0, for n >= 2.

T(2*n,n) = 2*A000108(n-1).

T(2*n+3,n) = -A120009(n).

T(n+10,3) = -A137742(n+2).

Sum_{k=0..n} T(n,k)*2^k = -(3^(n-2)) where n >= 2.

Sum_{k=0..n} T(n,k)*3^k = -(4^(n-1)) where n >= 2.

Sum_{k=0..n} T(n,k)*p^k = -((p+1)^(n-2))*((p-1)^2) where n >= 2.